Question

Find the equations of any two medians of the triangle formed by the lines 3x + 2y + 6 =0;
2x – 5y + 4 = 0 and x – 3y – 6 =0. Find also the centroid of the triangle​

Answer 1

Dear Student,

Let A, B and C be the three vertices of the triangle.
L₁: 3x + 2y + 6 = 0 , L₂ : 2x - 5y + 4 = 0 , L₃ : x - 3y - 6 = 0

A be the point of intersection of L₁ and L₂
3x + 2y = -6 ...................... (1)
2x - 5y = - 4 ...................... (2)

Multiplying (1) by 2 and (2) by 3 we get,
6x + 4y = - 12 ..................... (3)
6x - 15y = -12 ..................... (4)

Subtracting (3) from (4) we get,

- 15y - 4y = -12 - (-12)
- 19y = 0
y = 0

From (1) ,
3x + 2y = - 6
3x = - 6
x = - 2

Hence, A (-2 , 0)

B be the point of intersection of L₂ and L₃ :
Solving two lines we get, B (-42 , - 16)

Similarly, C be the intersection of L₁ and L₃ :
Solving two lines we get, C $\left(-\frac{6}{11},-\frac{24}{11}\right)$
Now, Let D be the mid point of BC

D $\left(\frac{-42-{\displaystyle \frac{6}{11}}}{2},\frac{-16-{\displaystyle \frac{24}{11}}}{2}\right)$ = D$\left(-\frac{234}{11},-\frac{100}{11}\right)$
Therefore, AD is the median of the triangle

Equation of AD is : y - 0 = $\frac{-{\displaystyle \frac{100}{11}}-0}{-{\displaystyle \frac{234}{11}}-\left(-2\right)}$ (x - (-2))
y = $\frac{25}{53}$ (x + 2)
53y = 25x + 50
25x - 53y + 50 = 0

Similarly equation for other medians can be found

Regards